If \(L\), \(M\), \(N\), and \(O\) are the mid-points of the sides of rectangle \(ABCD\). Prove that the quadrilateral \(LMNO\) formed by joining them is a rhombus.
Construction : Join \(A\) and \(C\)
Proof : Since \(ABCD\) is a rectangle, \(AD = BC\)
\(\Rightarrow \frac{1}{2}AD = \frac{1}{2}\)
\(\Rightarrow AO = BM\) and \(DO = CM\) - - - - - (I)
In \(\Delta ABC\), is the mid-point of \(AB\) and is the mid-point of \(BC\).
By the mid-point theorem 'The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is half of it.'
\(\Rightarrow LM\ || \ AC\) and \(LM = \frac{1}{2}\) - - - - (II)
Similarly, in \(\Delta ADC\), is the mid-point of \(CD\) and is the mid-point of \(AD\).
By the mid-point theorem,'The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is half of it.'
\(\Rightarrow NO \ || \ AC\) and \(NO= \frac{1}{2}\) - - - - (III)
From (I) and (II), we get, \(LM \ || \ NO\).
Thus, \(LMNO\) is a parallelogram.
In \(\Delta OAL\) and \(\Delta MBL\):
\(AL = \) [Given]
\(\angle OAL = \angle \) [Each \(90^\circ\)]
\(AO = BM\) [Using (I)]
Thus, \(\Delta OAL \cong \Delta MPL\) [by ].
\(\Rightarrow LO = LM\) [by CPCT]
But, we know that opposite sides of parallelogram are equal.
\(LM = NO\) and \(LO = NM\)
So, \(LM = NM = NO = LO\).
Hence, \(LMNO\) is a rhombus.
\(\Rightarrow \frac{1}{2}AD = \frac{1}{2}\)
\(\Rightarrow AO = BM\) and \(DO = CM\) - - - - - (I)
In \(\Delta ABC\), is the mid-point of \(AB\) and is the mid-point of \(BC\).
By the mid-point theorem 'The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is half of it.'
\(\Rightarrow LM\ || \ AC\) and \(LM = \frac{1}{2}\) - - - - (II)
Similarly, in \(\Delta ADC\), is the mid-point of \(CD\) and is the mid-point of \(AD\).
By the mid-point theorem,'The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is half of it.'
\(\Rightarrow NO \ || \ AC\) and \(NO= \frac{1}{2}\) - - - - (III)
From (I) and (II), we get, \(LM \ || \ NO\).
Thus, \(LMNO\) is a parallelogram.
In \(\Delta OAL\) and \(\Delta MBL\):
\(AL = \) [Given]
\(\angle OAL = \angle \) [Each \(90^\circ\)]
\(AO = BM\) [Using (I)]
Thus, \(\Delta OAL \cong \Delta MPL\) [by ].
\(\Rightarrow LO = LM\) [by CPCT]
But, we know that opposite sides of parallelogram are equal.
\(LM = NO\) and \(LO = NM\)
So, \(LM = NM = NO = LO\).
Hence, \(LMNO\) is a rhombus.